a) \(\sqrt{x}=2\)

\(\Rightarrow x=2^2=4\)

b) \(\sqrt{x-1-5}=\sqrt{x-6}=0\)

\(\Rightarrow x-6=0^2=0\)

\(\Rightarrow x=6\)

các câu sau tương tự

đăng ít ít thôi

Ta có: 2n+111...1 (n c/s 1)

xét 3 trường hợp: n=3k và n=3k+1; n=3k+2

TH1: Nếu n=3k thì 

*2n=6k =>2n chia hết cho 3 (1)

*Tổng các chữ số của 111...1(n c/s1) là 3k =>111...1 chia hết cho 3 (2)

Từ (1),(2) =>2n+111..1 chia hết cho 3

Tương tự với trường hợp 2 và 3 ta đều có 2n+111...1(n c/s 1) chia hết cho 3

Vậy.....

Wow

a) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow6006\left(x+1\right)+5460\left(x+1\right)+5005\left(x+1\right)=4620\left(x+1\right)+4290\left(x+1\right)\)

\(\Leftrightarrow\left(6006+5460+5005\right)\cdot\left(x+1\right)=\left(4620+4290\right)\cdot\left(x+1\right)\)

\(\Leftrightarrow16471\left(x+1\right)=8910\left(x+1\right)\)

\(\Leftrightarrow16471x+16471=8910x+8910\)

\(\Leftrightarrow16471x-8910x=8910-16471\)

\(\Leftrightarrow7561x=-7561\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

b) \(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)

\(\Rightarrow4096749040\left(x+4\right)+4094735904\left(x+3\right)=4092704785\left(x+2\right)+4090675680\left(x+1\right)\)

\(\Leftrightarrow4096769040x+16387076160+4094735904x+12284207712=4092704785x+8185409570+4090675680x+4090675680\)

\(\Leftrightarrow8191504944x+28671283872=8183380465x+12276085250\)

\(\Leftrightarrow8191504944x-8183380465x=12276085250-28671283872\)

\(\Leftrightarrow8124479x=-16395198622\)

\(\Rightarrow x=-2018\)

Vậy \(x=-2017\)

P/s: đây không phải cách làm tối ưu, vì vậy mình nghĩ bạn nên tham khảo từ các bài làm khác nhé!

4a)

\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2}+\sqrt{3}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2}-\sqrt{3}}\)

\(=\dfrac{2+\sqrt{3}}{2\sqrt{2}+\sqrt{3}}+\dfrac{2-\sqrt{3}}{-\sqrt{3}}\)

\(=\dfrac{\left(2+\sqrt{3}\right)\cdot\left(2\sqrt{2}-\sqrt{3}\right)}{5}-\dfrac{\left(2-\sqrt{3}\right)\sqrt{3}}{3}\)

\(=\dfrac{4\sqrt{2}-2\sqrt{3}+2\sqrt{6}-3}{5}-\dfrac{2\sqrt{3}-3}{3}\)

\(=\dfrac{3\left(4\sqrt{2}-2\sqrt{3}+2\sqrt{6}-3\right)-5\left(2\sqrt{3}-3\right)}{15}\)

\(=\dfrac{12\sqrt{2}-6\sqrt{3}+6\sqrt{6}-9-10\sqrt{3}+15}{15}\)

\(=\dfrac{12\sqrt{2}-16\sqrt{3}+6\sqrt{6}+6}{15}\)

a) \(\left|3x-5\right|-\left|x+2\right|=0\)

\(\Leftrightarrow\left|3x-5\right|=\left|x+2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=x+2\\3x-5=-\left(x+2\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{3}{4};x_2=\dfrac{7}{2}\)

b) \(\left|x\right|+\left|x+2\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+x+2=3\left(đk:x\ge0,x+2\ge0\right)\\-x+x+2=3\left(đk:x< 0,x+2\ge0\right)\\x-\left(x+2\right)=3\left(đk:x\ge0,x+2< 0\right)\\-x-\left(x+2\right)=3\left(đk:x< 0,x+2< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(đk:x\ge0,x\ge-2\right)\\x\in\varnothing\left(đk:x< 0,x\ge-2\right)\\x\in\varnothing\left(x\ge0,x< -2\right)\\x=-\dfrac{5}{2}\left(đk:x< 0,x< -2\right)\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{5}{2};x_2=\dfrac{1}{2}\)

đề sai